How to find all partitions of a set

Question

I have a set of distinct values. I am looking for a way to generate all partitions of this set, i.e. all possible ways of dividing the set into subsets.

For instance, the set {1, 2, 3} has the following partitions:

{ {1}, {2}, {3} },
{ {1, 2}, {3} },
{ {1, 3}, {2} },
{ {1}, {2, 3} },
{ {1, 2, 3} }.

As these are sets in the mathematical sense, order is irrelevant. For instance, {1, 2}, {3} is the same as {3}, {2, 1} and should not be a separate result.

A thorough definition of set partitions can be found on Wikipedia.

Answer 1

I've found a straightforward recursive solution.

First, let's solve a simpler problem: how to find all partitions consisting of exactly two parts. For an n-element set, we can count an int from 0 to (2^n)-1. This creates every n-bit pattern, with each bit corresponding to one input element. If the bit is 0, we place the element in the first part; if it is 1, the element is placed in the second part. This leaves one problem: For each partition, we'll get a duplicate result where the two parts are swapped. To remedy this, we'll always place the first element into the first part. We then only distribute the remaining n-1 elements by counting from 0 to (2^(n-1))-1.

Now that we can partition a set into two parts, we can write a recursive function that solves the rest of the problem. The function starts off with the original set and finds all two-part-partitions. For each of these partitions, it recursively finds all ways to partition the second part into two parts, yielding all three-part partitions. It then divides the last part of each of these partitions to generate all four-part partitions, and so on.

The following is an implementation in C#. Calling

Partitioning.GetAllPartitions(new[] { 1, 2, 3, 4 }) 

yields

{ {1, 2, 3, 4} }, { {1, 3, 4}, {2} }, { {1, 2, 4}, {3} }, { {1, 4}, {2, 3} }, { {1, 4}, {2}, {3} }, { {1, 2, 3}, {4} }, { {1, 3}, {2, 4} }, { {1, 3}, {2}, {4} }, { {1, 2}, {3, 4} }, { {1, 2}, {3}, {4} }, { {1}, {2, 3, 4} }, { {1}, {2, 4}, {3} }, { {1}, {2, 3}, {4} }, { {1}, {2}, {3, 4} }, { {1}, {2}, {3}, {4} }. 

using System; using System.Collections.Generic; using System.Linq;  namespace PartitionTest {     public static class Partitioning {         public static IEnumerable<T[][]> GetAllPartitions<T>(T[] elements) {             return GetAllPartitions(new T[][]{}, elements);         }          private static IEnumerable<T[][]> GetAllPartitions<T>(             T[][] fixedParts, T[] suffixElements)         {             // A trivial partition consists of the fixed parts             // followed by all suffix elements as one block             yield return fixedParts.Concat(new[] { suffixElements }).ToArray();              // Get all two-group-partitions of the suffix elements             // and sub-divide them recursively             var suffixPartitions = GetTuplePartitions(suffixElements);             foreach (Tuple<T[], T[]> suffixPartition in suffixPartitions) {                 var subPartitions = GetAllPartitions(                     fixedParts.Concat(new[] { suffixPartition.Item1 }).ToArray(),                     suffixPartition.Item2);                 foreach (var subPartition in subPartitions) {                     yield return subPartition;                 }             }         }          private static IEnumerable<Tuple<T[], T[]>> GetTuplePartitions<T>(             T[] elements)         {             // No result if less than 2 elements             if (elements.Length < 2) yield break;              // Generate all 2-part partitions             for (int pattern = 1; pattern < 1 << (elements.Length - 1); pattern++) {                 // Create the two result sets and                 // assign the first element to the first set                 List<T>[] resultSets = {                     new List<T> { elements[0] }, new List<T>() };                 // Distribute the remaining elements                 for (int index = 1; index < elements.Length; index++) {                     resultSets[(pattern >> (index - 1)) & 1].Add(elements[index]);                 }                  yield return Tuple.Create(                     resultSets[0].ToArray(), resultSets[1].ToArray());             }         }     } } 

Answer 2

Please refer to the Bell number, here is a brief thought to this problem:
consider f(n,m) as partition a set of n element into m non-empty sets.

For example, the partition of a set of 3 elements can be:
1) set size 1: {{1,2,3}, } <-- f(3,1)
2) set size 2: {{1,2},{3}}, {{1,3},{2}}, {{2,3},{1}} <-- f(3,2)
3) set size 3: {{1}, {2}, {3}} <-- f(3,3)

Now let's calculate f(4,2):
there are two ways to make f(4,2):

A. add a set to f(3,1), which will convert from {{1,2,3}, } to {{1,2,3}, {4}}
B. add 4 to any of set of f(3,2), which will convert from
{{1,2},{3}}, {{1,3},{2}}, {{2,3},{1}}
to
{{1,2,4},{3}}, {{1,2},{3,4}}
{{1,3,4},{2}}, {{1,3},{2,4}}
{{2,3,4},{1}}, {{2,3},{1,4}}

So f(4,2) = f(3,1) + f(3,2)*2
which result in f(n,m) = f(n-1,m-1) + f(n-1,m)*m

Here is Java code for get all partitions of set:

import java.util.ArrayList; import java.util.List;  public class SetPartition {     public static void main(String[] args) {         List<Integer> list = new ArrayList<>();         for(int i=1; i<=3; i++) {             list.add(i);         }          int cnt = 0;         for(int i=1; i<=list.size(); i++) {             List<List<List<Integer>>> ret = helper(list, i);             cnt += ret.size();             System.out.println(ret);         }         System.out.println("Number of partitions: " + cnt);     }      // partition f(n, m)     private static List<List<List<Integer>>> helper(List<Integer> ori, int m) {         List<List<List<Integer>>> ret = new ArrayList<>();         if(ori.size() < m || m < 1) return ret;          if(m == 1) {             List<List<Integer>> partition = new ArrayList<>();             partition.add(new ArrayList<>(ori));             ret.add(partition);             return ret;         }          // f(n-1, m)         List<List<List<Integer>>> prev1 = helper(ori.subList(0, ori.size() - 1), m);         for(int i=0; i<prev1.size(); i++) {             for(int j=0; j<prev1.get(i).size(); j++) {                 // Deep copy from prev1.get(i) to l                 List<List<Integer>> l = new ArrayList<>();                 for(List<Integer> inner : prev1.get(i)) {                     l.add(new ArrayList<>(inner));                 }                  l.get(j).add(ori.get(ori.size()-1));                 ret.add(l);             }         }          List<Integer> set = new ArrayList<>();         set.add(ori.get(ori.size() - 1));         // f(n-1, m-1)         List<List<List<Integer>>> prev2 = helper(ori.subList(0, ori.size() - 1), m - 1);         for(int i=0; i<prev2.size(); i++) {             List<List<Integer>> l = new ArrayList<>(prev2.get(i));             l.add(set);             ret.add(l);         }          return ret;     }  } 

And result is:
[[[1, 2, 3]]] [[[1, 3], [2]], [[1], [2, 3]], [[1, 2], [3]]] [[[1], [2], [3]]] Number of partitions: 5

Answer 3

Just for fun, here's a shorter purely iterative version:

public static IEnumerable<List<List<T>>> GetAllPartitions<T>(T[] elements) {
    var lists = new List<List<T>>();
    var indexes = new int[elements.Length];
    lists.Add(new List<T>());
    lists[0].AddRange(elements);
    for (;;) {
        yield return lists;
        int i,index;
        for (i=indexes.Length-1;; --i) {
            if (i<=0)
                yield break;
            index = indexes[i];
            lists[index].RemoveAt(lists[index].Count-1);
            if (lists[index].Count>0)
                break;
            lists.RemoveAt(index);
        }
        ++index;
        if (index >= lists.Count)
            lists.Add(new List<T>());
        for (;i<indexes.Length;++i) {
            indexes[i]=index;
            lists[index].Add(elements[i]);
            index=0;
        }
    }

Test here:https://ideone.com/EccB5n

And a simpler recursive version:

public static IEnumerable<List<List<T>>> GetAllPartitions<T>(T[] elements, int maxlen) {
    if (maxlen<=0) {
        yield return new List<List<T>>();
    }
    else {
        T elem = elements[maxlen-1];
        var shorter=GetAllPartitions(elements,maxlen-1);
        foreach (var part in shorter) {
            foreach (var list in part.ToArray()) {
                list.Add(elem);
                yield return part;
                list.RemoveAt(list.Count-1);
            }
            var newlist=new List<T>();
            newlist.Add(elem);
            part.Add(newlist);
            yield return part;
            part.RemoveAt(part.Count-1);
        }
    }

https://ideone.com/Kdir4e